因为b=y/a代入ab=a+b+3
得 a* (y/a)=a+(y/a)+3
==>y=a+y/a+3
两边同时乘于a
得ay=a^2+y+3a
将右边的y移到左边得
ay-y=a^2+3a
==>(a-1)y=a^2+3a
==>y=(a^2+3a)/(a-1)
一道高一函数中关于化简的问题若正数a,b满足ab=a+b+3,求ab的取值范围解法：设ab=y,则b=y/a,代入ab=
因为b=y/a代入ab=a+b+3
得 a* (y/a)=a+(y/a)+3
==>y=a+y/a+3
两边同时乘于a
得ay=a^2+y+3a
将右边的y移到左边得
ay-y=a^2+3a
==>(a-1)y=a^2+3a
==>y=(a^2+3a)/(a-1)